# L’Hôpital’s rule

Evaluating the limit of x-3 with x approaching 4 is very easy and straightforward: you just plug in the value x tends to and you’re done!

$\lim_{x\to4}\,&space;(x-3)=4-3=1$

Another example:

$\lim_{x\to\frac{\pi}{2}}\,&space;(\sin(x)+\cos(x))=\left&space;(&space;\sin\left&space;(&space;\frac{\pi}{2}&space;\right&space;)&space;+\cos\left&space;(&space;\frac{\pi}{2}&space;\right&space;)\right&space;)=1+0=1$

$\lim_{x\to&space;\infty}\frac{1}{x}=\frac{1}{\infty}=0$

As we saw here, one divided by zero equals infinity, therefore one divided by infinity equals zero.

We can’t evaluate limits directly by substitution when, if we do substitute, we get an indeterminate result, like infinity/infinity, 0/0, infinityinfinity, 0 times infinity, infinity/0. When this happens, we need to rewrite the limit, without substitution, and apply L’Hôpital’s rule. But what is it?

$\lim_{x\to&space;\frac{\pi}{2}}\left&space;(&space;\sec(x)&space;-\tan(x)\right&space;)$

I’ve come up with this limit because if we plug in pi/2 we get infinityinfinity, one of the indeterminate results, but we want to know the numeric result. Let’s see what we can do before we use L’Hôpital’s rule.

sec(x) can be written as 1/cos(x), and tan(x) as sin(x)/cos(x).

$\lim_{x\to&space;\frac{\pi}{2}}\left&space;(&space;\frac{1}{\cos(x)}&space;-\frac{\sin(x)}{\cos(x)}\right&space;)$

Let’s simplify this:

$\lim_{x\to&space;\frac{\pi}{2}}\left&space;(&space;\frac{1-\sin(x)}{\cos(x)}\right&space;)$

This is equivalent to the expression we had at the beginning, so if we use substitution we will still get an indeterminate result:

$\frac{1-\sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2})}=\frac{1-\sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2})}=\frac{1-1}{0}=\frac{0}{0}=\text{indeterminate}$

This means that infinityinfinity is equal to 0/0. But what’s the result?

L’Hôpital’s rule helps us solve these kind of limits by taking the derivative of nominator and denominator. Both must have at least a function of x, i.e. x must appear in both. If there’s no x present at the denominator, you can do this: let’s assume we have x/3; in order to apply the rule we can write it as x²/3x. As you may have noticed, the two expressions are equivalent.

Now, back to the limit we had:

$\lim_{x\to\frac{\pi}{2}}\frac{1-\sin(x)}{\cos(x)}$

Let’s apply the rule:

$\lim_{x\to\frac{\pi}{2}}\frac{\frac{\mathrm{d}&space;}{\mathrm{d}&space;x}\left&space;(&space;1-\sin(x)&space;\right&space;)}{\frac{\mathrm{d}&space;}{\mathrm{d}&space;x}\cos(x)}\rightarrow&space;\lim_{x\to\frac{\pi}{2}}\frac{-\cos(x)}{-\sin(x)}\rightarrow&space;\lim_{x\to\frac{\pi}{2}}\frac{\cos(x)}{\sin(x)}$

Notice how I used the right arrow and not the equal sign; this is because we still have a limit on the other sides, so the previous limit tends to (—>) the other, or at least this is what I think. If you wish, you can use the ‘=’ though.

Once we’re done with the rule, we can try and substitute the value x tends to:

$\lim_{x\to\frac{\pi}{2}}\frac{\cos(x)}{\sin(x)}=\frac{\cos\left&space;(&space;\frac{\pi}{2}&space;\right&space;)}{\sin\left&space;(&space;\frac{\pi}{2}&space;\right&space;)}=\frac{0}{1}=0$

Who would have imagined that infinityinfinity and 0/0 are equal to 0? Although that seemed pretty obvious already at the beginning, we had to prove it.

If you still get an indeterminate result after applying L’Hôpital’s rule you can apply it again, until you get the result.

I hope this explanation was clear, and if and if there’s something you didn’t quite understand, leave a comment down below and I will be happy to help you! Subscribe to stay tuned!

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